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Proof that a function with a countable set of discontinuities is Riemann integrable without the notion of...
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Let $f:[a,b]to mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion).
One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $lambda(A)=0$
EDIT: Here is the proof of WimC with all the details:
Let $epsilon>0$ and $$D=left{d_1,d_2,...right}subseteq A$$ be the countable set of discontinuities of $f$. Define:
$$I=left{xin [a,b]:exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsilonright}$$
Now
$$xin Iiff exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsiloniff [left|y-xright|<deltaimplies omega f(y)<epsilon]$$
and because $epsilon$ is in fact arbitrary,
$$xin Iiff text{ $f$ is continuous at $x$}$$
In addition, if $xin I$, $exists delta>0$ so that
$$omega f((x-delta,x+delta)cap [a,b])<epsilon$$
If $yin B(x,delta)cap [a,b]$. Then $yin I$ and so $I$ is open relative to $[a,b]$.
Because $I=[a,b]setminus D$, $[a,b]=Icup D$. For $kin mathbb{N}$ define
$$D_k=left(d_k-frac{epsilon}{M2^{k+1}},d_k+frac{epsilon}{M2^{k+1}}right)cap [a,b]$$
Obviously $Dsubset bigcup_{k=1}^{infty}D_k$ and $[a,b]=Icup bigcup_{k=1}^{infty}D_k$ (since $D_ksubseteq [a,b]$). The compactness of $[a,b]$ implies $[a,b]=Icup bigcup_{k=1}^{N}D_k$.
Now $[a,b]setminus bigcup_{k=1}^{N}D_k$ is compact (closed and bounded) and included in $I$.
As such it can be covered by
$$F_x=(x-delta_x,x+delta_x)$$
where $ delta_x>0:$ is chosen so that $omega f((x-delta_x,x+delta_x)cap [a,b])<epsilon$.
Compactness implies the existence of a finite subcover,
$$[a,b]setminus bigcup_{k=1}^{N}D_ksubseteq bigcup_{i=1}^{M}(x_i-delta_i,x+delta_i)$$
As we can replace the intervals that intersect we can suppose
$$ bigcap_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Therefore,
$$[a,b]= bigcup_{k=1}^{N}overline{D}_kcup bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=bigcup_{i=0}^{n}[t_{i-1},t_i]$$
where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Considering all the endpoints of the above (that are pairwise different) we can create a partition $mathcal{P}=left{a=t_0<...<t_n=bright}$ of $[a,b]$. We separate the indices: $A=left{i:[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]right}$ and $B=left{i:[t_{i-1},t_i]= overline{D}_k right}$. Therefore,
begin{gather}U_{f,mathcal{P}}-L_{f,mathcal{P}}=sum_{i=1}^nomega f([t_{i-1},t_i])(t_i-t_{i-1})=sum_{iin A}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])+sum_{iin B}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])\
le sum_{iin A}2left|fright|ell([t_{i-1},t_i])+sum_{iin B}epsilonell(overline{D}_k)le 2left|fright|frac{epsilon}{M}+epsilon(b-a)=2epsilon+epsilon(b-a)end{gather}
My questions are: Is this proof correct? ( I doubt the point:
"where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$")
Second, can it be simplified?
calculus real-analysis integration
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
add a comment |
up vote
16
down vote
favorite
Let $f:[a,b]to mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion).
One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $lambda(A)=0$
EDIT: Here is the proof of WimC with all the details:
Let $epsilon>0$ and $$D=left{d_1,d_2,...right}subseteq A$$ be the countable set of discontinuities of $f$. Define:
$$I=left{xin [a,b]:exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsilonright}$$
Now
$$xin Iiff exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsiloniff [left|y-xright|<deltaimplies omega f(y)<epsilon]$$
and because $epsilon$ is in fact arbitrary,
$$xin Iiff text{ $f$ is continuous at $x$}$$
In addition, if $xin I$, $exists delta>0$ so that
$$omega f((x-delta,x+delta)cap [a,b])<epsilon$$
If $yin B(x,delta)cap [a,b]$. Then $yin I$ and so $I$ is open relative to $[a,b]$.
Because $I=[a,b]setminus D$, $[a,b]=Icup D$. For $kin mathbb{N}$ define
$$D_k=left(d_k-frac{epsilon}{M2^{k+1}},d_k+frac{epsilon}{M2^{k+1}}right)cap [a,b]$$
Obviously $Dsubset bigcup_{k=1}^{infty}D_k$ and $[a,b]=Icup bigcup_{k=1}^{infty}D_k$ (since $D_ksubseteq [a,b]$). The compactness of $[a,b]$ implies $[a,b]=Icup bigcup_{k=1}^{N}D_k$.
Now $[a,b]setminus bigcup_{k=1}^{N}D_k$ is compact (closed and bounded) and included in $I$.
As such it can be covered by
$$F_x=(x-delta_x,x+delta_x)$$
where $ delta_x>0:$ is chosen so that $omega f((x-delta_x,x+delta_x)cap [a,b])<epsilon$.
Compactness implies the existence of a finite subcover,
$$[a,b]setminus bigcup_{k=1}^{N}D_ksubseteq bigcup_{i=1}^{M}(x_i-delta_i,x+delta_i)$$
As we can replace the intervals that intersect we can suppose
$$ bigcap_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Therefore,
$$[a,b]= bigcup_{k=1}^{N}overline{D}_kcup bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=bigcup_{i=0}^{n}[t_{i-1},t_i]$$
where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Considering all the endpoints of the above (that are pairwise different) we can create a partition $mathcal{P}=left{a=t_0<...<t_n=bright}$ of $[a,b]$. We separate the indices: $A=left{i:[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]right}$ and $B=left{i:[t_{i-1},t_i]= overline{D}_k right}$. Therefore,
begin{gather}U_{f,mathcal{P}}-L_{f,mathcal{P}}=sum_{i=1}^nomega f([t_{i-1},t_i])(t_i-t_{i-1})=sum_{iin A}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])+sum_{iin B}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])\
le sum_{iin A}2left|fright|ell([t_{i-1},t_i])+sum_{iin B}epsilonell(overline{D}_k)le 2left|fright|frac{epsilon}{M}+epsilon(b-a)=2epsilon+epsilon(b-a)end{gather}
My questions are: Is this proof correct? ( I doubt the point:
"where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$")
Second, can it be simplified?
calculus real-analysis integration
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
1
Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05
add a comment |
up vote
16
down vote
favorite
up vote
16
down vote
favorite
Let $f:[a,b]to mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion).
One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $lambda(A)=0$
EDIT: Here is the proof of WimC with all the details:
Let $epsilon>0$ and $$D=left{d_1,d_2,...right}subseteq A$$ be the countable set of discontinuities of $f$. Define:
$$I=left{xin [a,b]:exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsilonright}$$
Now
$$xin Iiff exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsiloniff [left|y-xright|<deltaimplies omega f(y)<epsilon]$$
and because $epsilon$ is in fact arbitrary,
$$xin Iiff text{ $f$ is continuous at $x$}$$
In addition, if $xin I$, $exists delta>0$ so that
$$omega f((x-delta,x+delta)cap [a,b])<epsilon$$
If $yin B(x,delta)cap [a,b]$. Then $yin I$ and so $I$ is open relative to $[a,b]$.
Because $I=[a,b]setminus D$, $[a,b]=Icup D$. For $kin mathbb{N}$ define
$$D_k=left(d_k-frac{epsilon}{M2^{k+1}},d_k+frac{epsilon}{M2^{k+1}}right)cap [a,b]$$
Obviously $Dsubset bigcup_{k=1}^{infty}D_k$ and $[a,b]=Icup bigcup_{k=1}^{infty}D_k$ (since $D_ksubseteq [a,b]$). The compactness of $[a,b]$ implies $[a,b]=Icup bigcup_{k=1}^{N}D_k$.
Now $[a,b]setminus bigcup_{k=1}^{N}D_k$ is compact (closed and bounded) and included in $I$.
As such it can be covered by
$$F_x=(x-delta_x,x+delta_x)$$
where $ delta_x>0:$ is chosen so that $omega f((x-delta_x,x+delta_x)cap [a,b])<epsilon$.
Compactness implies the existence of a finite subcover,
$$[a,b]setminus bigcup_{k=1}^{N}D_ksubseteq bigcup_{i=1}^{M}(x_i-delta_i,x+delta_i)$$
As we can replace the intervals that intersect we can suppose
$$ bigcap_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Therefore,
$$[a,b]= bigcup_{k=1}^{N}overline{D}_kcup bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=bigcup_{i=0}^{n}[t_{i-1},t_i]$$
where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Considering all the endpoints of the above (that are pairwise different) we can create a partition $mathcal{P}=left{a=t_0<...<t_n=bright}$ of $[a,b]$. We separate the indices: $A=left{i:[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]right}$ and $B=left{i:[t_{i-1},t_i]= overline{D}_k right}$. Therefore,
begin{gather}U_{f,mathcal{P}}-L_{f,mathcal{P}}=sum_{i=1}^nomega f([t_{i-1},t_i])(t_i-t_{i-1})=sum_{iin A}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])+sum_{iin B}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])\
le sum_{iin A}2left|fright|ell([t_{i-1},t_i])+sum_{iin B}epsilonell(overline{D}_k)le 2left|fright|frac{epsilon}{M}+epsilon(b-a)=2epsilon+epsilon(b-a)end{gather}
My questions are: Is this proof correct? ( I doubt the point:
"where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$")
Second, can it be simplified?
calculus real-analysis integration
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
Let $f:[a,b]to mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion).
One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $lambda(A)=0$
EDIT: Here is the proof of WimC with all the details:
Let $epsilon>0$ and $$D=left{d_1,d_2,...right}subseteq A$$ be the countable set of discontinuities of $f$. Define:
$$I=left{xin [a,b]:exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsilonright}$$
Now
$$xin Iiff exists delta>0: omega f((x-delta,x+delta)cap [a,b])<epsiloniff [left|y-xright|<deltaimplies omega f(y)<epsilon]$$
and because $epsilon$ is in fact arbitrary,
$$xin Iiff text{ $f$ is continuous at $x$}$$
In addition, if $xin I$, $exists delta>0$ so that
$$omega f((x-delta,x+delta)cap [a,b])<epsilon$$
If $yin B(x,delta)cap [a,b]$. Then $yin I$ and so $I$ is open relative to $[a,b]$.
Because $I=[a,b]setminus D$, $[a,b]=Icup D$. For $kin mathbb{N}$ define
$$D_k=left(d_k-frac{epsilon}{M2^{k+1}},d_k+frac{epsilon}{M2^{k+1}}right)cap [a,b]$$
Obviously $Dsubset bigcup_{k=1}^{infty}D_k$ and $[a,b]=Icup bigcup_{k=1}^{infty}D_k$ (since $D_ksubseteq [a,b]$). The compactness of $[a,b]$ implies $[a,b]=Icup bigcup_{k=1}^{N}D_k$.
Now $[a,b]setminus bigcup_{k=1}^{N}D_k$ is compact (closed and bounded) and included in $I$.
As such it can be covered by
$$F_x=(x-delta_x,x+delta_x)$$
where $ delta_x>0:$ is chosen so that $omega f((x-delta_x,x+delta_x)cap [a,b])<epsilon$.
Compactness implies the existence of a finite subcover,
$$[a,b]setminus bigcup_{k=1}^{N}D_ksubseteq bigcup_{i=1}^{M}(x_i-delta_i,x+delta_i)$$
As we can replace the intervals that intersect we can suppose
$$ bigcap_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Therefore,
$$[a,b]= bigcup_{k=1}^{N}overline{D}_kcup bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=bigcup_{i=0}^{n}[t_{i-1},t_i]$$
where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$
Considering all the endpoints of the above (that are pairwise different) we can create a partition $mathcal{P}=left{a=t_0<...<t_n=bright}$ of $[a,b]$. We separate the indices: $A=left{i:[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]right}$ and $B=left{i:[t_{i-1},t_i]= overline{D}_k right}$. Therefore,
begin{gather}U_{f,mathcal{P}}-L_{f,mathcal{P}}=sum_{i=1}^nomega f([t_{i-1},t_i])(t_i-t_{i-1})=sum_{iin A}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])+sum_{iin B}omega f([t_{i-1},t_i])ell([t_{i-1},t_i])\
le sum_{iin A}2left|fright|ell([t_{i-1},t_i])+sum_{iin B}epsilonell(overline{D}_k)le 2left|fright|frac{epsilon}{M}+epsilon(b-a)=2epsilon+epsilon(b-a)end{gather}
My questions are: Is this proof correct? ( I doubt the point:
"where for $ile n$, $[t_{i-1},t_i]= [x_k-delta_k,x+delta_k]$ or $[t_{i-1},t_i]= overline{D}_k$ because
$$bigcup_{k=1}^{N}overline{D}_kcap bigcup_{i=1}^{M}[x_i-delta_i,x+delta_i]=emptyset$$")
Second, can it be simplified?
calculus real-analysis integration
calculus real-analysis integration
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
edited Dec 26 '12 at 8:01
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
asked Dec 21 '12 at 12:17
Nameless
10.3k12055
10.3k12055
1
Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05
add a comment |
1
Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05
1
1
Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05
add a comment |
1 Answer
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Let $M = sup(f) - inf(f)$ on $[a,b]$ and $D = {d_0, d_1, dotsc } subset [a,b]$ be the countable set of discontinuities of $f$. Let $delta > 0$ and define $I subseteq [a,b]$ as $$I = { x in [a,b] mid sup(f) - inf(f) < frac{delta}{b-a} textrm{ on some neighbourhood of }x }.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I cup D$. For $k geq 0$ define the interval $D_k$ by $$D_k = left(d_k-frac{delta}{M2^{k+2}}, , d_k+frac{delta}{M2^{k+2}} right) cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I cup D_0 cup D_1 cup dotsc cup D_n$$ for some $n geq 0$. The complement $$[a,b] setminus (D_0 cup dotsc cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $sup(f)-inf(f) < delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $sup(f) - inf(f) leq delta/(b-a)$ on each. Finally the closure of $D_0 cup dotsc cup D_n$ is itself a union of closed intervals with a total length less than $delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most
$2delta$. Since $delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.
answered Dec 23 '12 at 21:27
WimC
24k22961
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
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Let $M = sup(f) - inf(f)$ on $[a,b]$ and $D = {d_0, d_1, dotsc } subset [a,b]$ be the countable set of discontinuities of $f$. Let $delta > 0$ and define $I subseteq [a,b]$ as $$I = { x in [a,b] mid sup(f) - inf(f) < frac{delta}{b-a} textrm{ on some neighbourhood of }x }.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I cup D$. For $k geq 0$ define the interval $D_k$ by $$D_k = left(d_k-frac{delta}{M2^{k+2}}, , d_k+frac{delta}{M2^{k+2}} right) cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I cup D_0 cup D_1 cup dotsc cup D_n$$ for some $n geq 0$. The complement $$[a,b] setminus (D_0 cup dotsc cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $sup(f)-inf(f) < delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $sup(f) - inf(f) leq delta/(b-a)$ on each. Finally the closure of $D_0 cup dotsc cup D_n$ is itself a union of closed intervals with a total length less than $delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most
$2delta$. Since $delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.
answered Dec 23 '12 at 21:27
WimC
24k22961
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
|
show 7 more comments
up vote
10
down vote
accepted
Let $M = sup(f) - inf(f)$ on $[a,b]$ and $D = {d_0, d_1, dotsc } subset [a,b]$ be the countable set of discontinuities of $f$. Let $delta > 0$ and define $I subseteq [a,b]$ as $$I = { x in [a,b] mid sup(f) - inf(f) < frac{delta}{b-a} textrm{ on some neighbourhood of }x }.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I cup D$. For $k geq 0$ define the interval $D_k$ by $$D_k = left(d_k-frac{delta}{M2^{k+2}}, , d_k+frac{delta}{M2^{k+2}} right) cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I cup D_0 cup D_1 cup dotsc cup D_n$$ for some $n geq 0$. The complement $$[a,b] setminus (D_0 cup dotsc cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $sup(f)-inf(f) < delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $sup(f) - inf(f) leq delta/(b-a)$ on each. Finally the closure of $D_0 cup dotsc cup D_n$ is itself a union of closed intervals with a total length less than $delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most
$2delta$. Since $delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.
answered Dec 23 '12 at 21:27
WimC
24k22961
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
|
show 7 more comments
up vote
10
down vote
accepted
up vote
10
down vote
accepted
Let $M = sup(f) - inf(f)$ on $[a,b]$ and $D = {d_0, d_1, dotsc } subset [a,b]$ be the countable set of discontinuities of $f$. Let $delta > 0$ and define $I subseteq [a,b]$ as $$I = { x in [a,b] mid sup(f) - inf(f) < frac{delta}{b-a} textrm{ on some neighbourhood of }x }.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I cup D$. For $k geq 0$ define the interval $D_k$ by $$D_k = left(d_k-frac{delta}{M2^{k+2}}, , d_k+frac{delta}{M2^{k+2}} right) cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I cup D_0 cup D_1 cup dotsc cup D_n$$ for some $n geq 0$. The complement $$[a,b] setminus (D_0 cup dotsc cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $sup(f)-inf(f) < delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $sup(f) - inf(f) leq delta/(b-a)$ on each. Finally the closure of $D_0 cup dotsc cup D_n$ is itself a union of closed intervals with a total length less than $delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most
$2delta$. Since $delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.
answered Dec 23 '12 at 21:27
WimC
24k22961
Let $M = sup(f) - inf(f)$ on $[a,b]$ and $D = {d_0, d_1, dotsc } subset [a,b]$ be the countable set of discontinuities of $f$. Let $delta > 0$ and define $I subseteq [a,b]$ as $$I = { x in [a,b] mid sup(f) - inf(f) < frac{delta}{b-a} textrm{ on some neighbourhood of }x }.$$ Then $I$ is open and contains all points where $f$ is continuous. In particular $[a,b] = I cup D$. For $k geq 0$ define the interval $D_k$ by $$D_k = left(d_k-frac{delta}{M2^{k+2}}, , d_k+frac{delta}{M2^{k+2}} right) cap [a,b].$$ Then $I$ together with all these intervals cover $[a,b]$ and because $[a,b]$ is compact it is already covered by a finite union $$[a,b] = I cup D_0 cup D_1 cup dotsc cup D_n$$ for some $n geq 0$. The complement $$[a,b] setminus (D_0 cup dotsc cup D_n)$$ is compact and contained in $I$ and can therefore be covered by open intervals such that $sup(f)-inf(f) < delta/(b-a)$ on the closure of each. (Every point in $I$ has such a neighbourhood by definition.) By compactness a finite subcover of such intervals exists. Considering all end points in this subcover as possible cut points one can partition the complement into finitely many closed intervals such that $sup(f) - inf(f) leq delta/(b-a)$ on each. Finally the closure of $D_0 cup dotsc cup D_n$ is itself a union of closed intervals with a total length less than $delta/M$.
The entire interval $[a,b]$ is now partitioned in a finite number of closed intervals that admit upper and lower Riemann sums of $f$ that differ by at most
$2delta$. Since $delta$ can be chosen arbitrarily small, $f$ is Riemann integrable.
In fact this argument works just as well for any $D$ that can be covered by a countable union of intervals of arbitrarily small total length, i.e. for $D$ of measure $0$.
answered Dec 23 '12 at 21:27
WimC
24k22961
edited Dec 25 '12 at 19:07
answered Dec 23 '12 at 21:27
WimC
24k22961
answered Dec 23 '12 at 21:27
WimC
24k22961
answered Dec 23 '12 at 21:27
WimC
24k22961
24k22961
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
|
show 7 more comments
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
First thank you for a great answer. I have some questions: 1. In the definition of $I$, I suppose $f$ is restricted in that neiborhood of $x$ and the supremum,infimum are taken there. Is that correct?
– Nameless
Dec 24 '12 at 7:52
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
2. "Therefore it allows a finite partition in closed intervals". Can you make this part clearer by defining the partition?
– Nameless
Dec 24 '12 at 8:13
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
@Nameless Yes $f$ is of course restricted to a neighbourhood in the definition of $I$. I can add more details to the "finite partition part" in the answer but it may have to wait until after christmas. But here's the idea: every point in $I$ is by definition contained in some open interval in $I$ on which $sup - inf$ is sufficiently small. Since $I setminus D_0 cup dotsc cup D_n$ is compact, it is covered by finitely many such intervals. Now basically take all the end points of these intervals as cut points for the partition.
– WimC
Dec 24 '12 at 8:54
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
I see. I will try and fill all the missing details in your proof. If I am successful I will notify you after christmas.
– Nameless
Dec 24 '12 at 8:57
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
Well I have reread your answer and understand everything up to the point: "Therefore it allows..." Could add more details from then on?
– Nameless
Dec 25 '12 at 18:39
|
show 7 more comments
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Do you mean Riemann integrable?
– Quinn Culver
Dec 22 '12 at 21:58
@QuinnCulver Yes.
– Nameless
Dec 22 '12 at 21:59
@Nameless, I think there is a typo in your edit. Set $I$ is constructed to depend on $epsilon$ so $I$ contains all points of continuity, but it can contain points of discontinuity as well.
– jdods
Apr 20 at 8:05